Aldehydes, Ketones, & Carboxylic Acids
Name Reactions, Reaction Mechanisms, Acidity Trends & Fully Solved Board PYQs
1. Structure and Nature of Carbonyl Group
Aldehydes ($-CHO$), ketones ($>C=O$), and carboxylic acids ($-COOH$) all contain the carbonyl group ($>C=O$).
The carbon atom of the carbonyl group is $sp^2$ hybridized and forms three sigma ($\sigma$) bonds. The unhybridized p-orbital overlaps with the p-orbital of oxygen to form a pi ($\pi$) bond. The geometry is trigonal planar with bond angles of ~120°.
Polarity of Carbonyl Group
Oxygen is highly electronegative compared to carbon. Thus, the $\pi$ electrons are pulled towards oxygen, making it partially negative ($\delta^-$) and the carbon partially positive ($\delta^+$).
$>C^{\delta+} = O^{\delta-}$
Because of this electrophilic carbon ($\delta^+$), carbonyl compounds readily undergo Nucleophilic Addition Reactions.
2. Preparation of Aldehydes and Ketones (Name Reactions)
1. Rosenmund Reduction (Only for Aldehydes)
Acyl chlorides (acid chlorides) are reduced to aldehydes by passing hydrogen gas over a catalyst composed of Palladium ($Pd$) supported on Barium Sulfate ($BaSO_4$) poisoned with sulfur or quinoline. The poison prevents further reduction of aldehyde to alcohol.
$R-CO-Cl + H_2 \xrightarrow{Pd/BaSO_4 \text{ (quinoline)}} R-CHO + HCl$
Example: Acetyl chloride gives Acetaldehyde. (Note: Formaldehyde cannot be prepared this way as formyl chloride is highly unstable).
2. Stephen Reaction (From Nitriles)
Alkyl nitriles ($R-CN$) are reduced to corresponding imine hydrochlorides using Stannous chloride ($SnCl_2$) and concentrated $HCl$. This upon acid hydrolysis gives aldehydes.
$R-C \equiv N + 2[H] \xrightarrow{SnCl_2 / HCl} R-CH=NH\cdot HCl$
$R-CH=NH\cdot HCl + H_2O \xrightarrow{\Delta} R-CHO + NH_4Cl$
3. Etard Reaction (Preparation of Benzaldehyde)
Toluene is oxidized to benzaldehyde using Chromyl chloride ($CrO_2Cl_2$) in a non-polar solvent like $CS_2$. A brown chromium complex is formed, which on acid hydrolysis yields benzaldehyde.
$C_6H_5-CH_3 + 2CrO_2Cl_2 \xrightarrow{CS_2} C_6H_5-CH(OCrOHCl_2)_2$
$C_6H_5-CH(OCrOHCl_2)_2 \xrightarrow{H_3O^+} C_6H_5CHO \text{ (Benzaldehyde)}$
4. Gatterman-Koch Formylation
Benzene is treated with a mixture of carbon monoxide and hydrogen chloride gas under high pressure in the presence of anhydrous $AlCl_3$ and cuprous chloride to form Benzaldehyde.
$C_6H_6 + CO + HCl \xrightarrow{\text{anh. } AlCl_3 / CuCl} C_6H_5CHO + HCl$
5. Friedel-Crafts Acylation (For Ketones)
Benzene reacts with acyl chlorides or acid anhydrides in the presence of anhydrous $AlCl_3$ to yield aromatic ketones.
$C_6H_6 + CH_3COCl \xrightarrow{\text{anh. } AlCl_3} C_6H_5COCH_3 \text{ (Acetophenone)} + HCl$
3. Preparation of Carboxylic Acids
- From Primary Alcohols & Aldehydes: Readily oxidized to carboxylic acids using strong oxidizing agents like Acidified Potassium dichromate ($K_2Cr_2O_7 / H_2SO_4$) or alkaline $KMnO_4$.
- From Nitriles (Alkyl Cyanides): Nitriles undergo complete hydrolysis on boiling with dilute mineral acids or alkalies to form carboxylic acids.
$R-C \equiv N + 2H_2O + HCl \xrightarrow{\Delta} R-COOH + NH_4Cl$ - From Grignard Reagent: $RMgX$ reacts with solid carbon dioxide (Dry Ice) in dry ether to form an adduct which on acid hydrolysis yields carboxylic acid.
$R-Mg-X + O=C=O \xrightarrow{\text{dry ether}} R-COOMgX \xrightarrow{H_3O^+} R-COOH$ - From Alkylbenzenes: Aromatic carboxylic acids can be prepared by vigorous oxidation of alkylbenzenes (like Toluene) with acidic or alkaline $KMnO_4$. The entire alkyl side chain (regardless of length) is oxidized to a $-COOH$ group.
$C_6H_5CH_3 \xrightarrow{KMnO_4 / KOH, \Delta} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH \text{ (Benzoic acid)}$
4. Chemical Reactions of Aldehydes and Ketones
A. Reactivity in Nucleophilic Addition
Aldehydes are more reactive than ketones towards nucleophilic addition due to:
- Steric Reason: Ketones have two bulky alkyl groups that hinder the approach of the nucleophile. Aldehydes have only one alkyl group and a small H atom.
- Electronic Reason: The two alkyl groups in ketones release electrons (+I effect), which reduces the positive charge ($\delta^+$) on the carbonyl carbon more effectively than the single alkyl group in aldehydes, thus reducing its electrophilicity.
B. Aldol Condensation (Requires $\alpha$-Hydrogen)
Aldehydes and ketones having at least one $\alpha$-hydrogen atom undergo self-addition in the presence of dilute alkali (like dil. $NaOH$ or $Ba(OH)_2$) to form $\beta$-hydroxy aldehydes (Aldol) or $\beta$-hydroxy ketones (Ketol). Upon heating, they lose a water molecule to form $\alpha,\beta$-unsaturated carbonyl compounds.
Reaction (Ethanal):
$2CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO \text{ (Acetaldol)}$
$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO \text{ (Crotonaldehyde)}$
C. Cannizzaro Reaction (No $\alpha$-Hydrogen)
Aldehydes which do not contain an $\alpha$-hydrogen atom (e.g., Formaldehyde $HCHO$, Benzaldehyde $C_6H_5CHO$) undergo self-oxidation-reduction (disproportionation) when heated with concentrated alkali (50% NaOH or KOH).
One molecule is reduced to alcohol while the other is oxidized to the salt of carboxylic acid.
Reaction (Formaldehyde):
$2HCHO + \text{conc. } NaOH \xrightarrow{\Delta} CH_3OH \text{ (Methanol)} + HCOONa \text{ (Sod. formate)}$
D. Reduction to Hydrocarbons
Clemmensen Reduction
Carbonyl group is reduced to $-CH_2-$ group by using Zinc amalgam ($Zn-Hg$) and concentrated $HCl$.
$>C=O + 4[H] \xrightarrow{Zn-Hg / \text{conc. } HCl} >CH_2 + H_2O$
Wolff-Kishner Reduction
Carbonyl group is reduced to $-CH_2-$ by heating with Hydrazine ($NH_2NH_2$) and KOH in a high boiling solvent like ethylene glycol.
$>C=O \xrightarrow{NH_2NH_2} >C=N-NH_2 \xrightarrow{KOH/\text{Glycol}, \Delta} >CH_2 + N_2$
E. Oxidation (Tollens' and Fehling's Tests)
These tests distinguish Aldehydes from Ketones. Aldehydes are easily oxidized, while ketones are not.
- Tollens' Test (Silver Mirror Test): Ammoniacal silver nitrate solution. Aldehydes reduce $Ag^+$ to metallic silver ($Ag$), which deposits as a mirror on the test tube wall. Both aliphatic and aromatic aldehydes give this test.
- Fehling's Test: A mixture of Fehling A (aq. $CuSO_4$) and Fehling B (Rochelle salt + NaOH). Aliphatic aldehydes reduce $Cu^{2+}$ to red precipitate of cuprous oxide ($Cu_2O$). (Aromatic aldehydes do NOT give this test).
5. Chemical Reactions of Carboxylic Acids
Acidity of Carboxylic Acids
Carboxylic acids are the most acidic among common organic compounds (more acidic than phenols and alcohols) because the carboxylate ion ($RCOO^-$) formed after the loss of $H^+$ is highly resonance stabilized by two equivalent electronegative oxygen atoms.
Effect of Substituents on Acidity:
- Electron Withdrawing Groups (EWG): (like $-Cl, -NO_2, -F$). They exert a -I effect, dispersing the negative charge and stabilizing the carboxylate ion, thereby increasing acidic strength.
Order: $CCl_3COOH > CHCl_2COOH > CH_2ClCOOH > CH_3COOH$ - Electron Donating Groups (EDG): (like alkyl groups). They exert a +I effect, intensifying the negative charge and destabilizing the carboxylate ion, thereby decreasing acidic strength.
Esterification
Carboxylic acids react with alcohols or phenols in the presence of a strong acid catalyst (conc. $H_2SO_4$ or $HCl$ gas) to form esters. It is a reversible reaction.
$R-COOH + R'-OH \rightleftharpoons R-COOR' (\text{Ester}) + H_2O$
Hell-Volhard-Zelinsky (HVZ) Reaction
Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus to give $\alpha$-halocarboxylic acids.
$CH_3-COOH + Cl_2 \xrightarrow{\text{Red } P / H_2O} Cl-CH_2-COOH + HCl$
6. Solved Textbook Reasoning Questions
Question 1: Absence of Aldol Condensation
Why does benzaldehyde not undergo Aldol condensation?
Answer:
Aldol condensation specifically requires the presence of an $\alpha$-hydrogen atom in the carbonyl compound. In benzaldehyde ($C_6H_5CHO$), the carbonyl carbon is attached directly to the benzene ring. The carbon of the benzene ring attached to the $-CHO$ group (the $\alpha$-carbon) has no hydrogen atom attached to it. Due to the absence of $\alpha$-hydrogen, benzaldehyde cannot form the required enolate ion and therefore does not undergo Aldol condensation (instead, it undergoes Cannizzaro reaction with strong alkali).
Question 2: Acidity Comparison
Chloroacetic acid is stronger than acetic acid. Explain.
Answer:
Chloroacetic acid ($Cl-CH_2-COOH$) contains a highly electronegative Chlorine atom. Chlorine exerts a strong electron-withdrawing inductive effect (-I effect). This pulls the electron density away from the $O-H$ bond, facilitating the easy release of $H^+$. Furthermore, it stabilizes the resulting chloroacetate ion by dispersing the negative charge. Acetic acid ($CH_3-COOH$) contains a methyl group which exerts an electron-donating effect (+I effect), making the release of $H^+$ harder and destabilizing the acetate ion. Hence, chloroacetic acid is stronger.
7. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the IUPAC name of acetone. (Oct 2013, March 2017)
Answer: Propan-2-one.
Q2. Name the reagent used in Rosenmund reduction. (March 2016, Oct 2021)
Answer: Hydrogen gas ($H_2$) in the presence of Palladium on Barium sulfate ($Pd/BaSO_4$) poisoned with quinoline or sulfur.
Q3. Identify the product obtained when benzaldehyde reacts with Tollens' reagent. (March 2018)
Answer: Benzoate ion (or Benzoic acid on acidification) and a silver mirror ($Ag$).
2 Mark Questions (SA-I)
Q4. What is the action of hydrazine and KOH (Wolff-Kishner reduction) on acetaldehyde? Write the equation. (March 2014, March 2019)
Acetaldehyde is reduced to ethane. It first forms a hydrazone, which upon heating with KOH in ethylene glycol decomposes to give ethane.
$CH_3CHO + NH_2NH_2 \rightarrow CH_3-CH=N-NH_2 \xrightarrow{KOH, \Delta} CH_3-CH_3 + N_2 \uparrow$
Q5. Write a note on Stephen reaction. (Oct 2015, March 2022)
Alkyl nitriles on reduction with stannous chloride ($SnCl_2$) and concentrated hydrochloric acid ($HCl$) give imine hydrochlorides, which on acid hydrolysis yield corresponding aldehydes. This is called Stephen reaction.
$R-CN \xrightarrow{SnCl_2 / HCl} R-CH=NH\cdot HCl \xrightarrow{H_3O^+} R-CHO + NH_4Cl$
3/4 Mark Questions (LA)
Q6. Explain Aldol Condensation with a suitable example. State the condition necessary for this reaction. (Oct 2014, March 2019, March 2023)
Condition:
The aldehyde or ketone must contain at least one $\alpha$-hydrogen atom.
Explanation & Example:
When two molecules of an aldehyde containing an $\alpha$-hydrogen (e.g., acetaldehyde) are treated with dilute alkali (like dil. NaOH), they condense to form a $\beta$-hydroxy aldehyde (aldol). Upon heating, aldol readily loses a water molecule to form an $\alpha,\beta$-unsaturated aldehyde.
1. $2CH_3-CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \text{ (3-Hydroxybutanal)}$
2. $CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO \text{ (But-2-enal)}$
Q7. What is Cannizzaro reaction? Write the equation for the reaction of formaldehyde with concentrated NaOH. Why does formaldehyde undergo Cannizzaro reaction but acetaldehyde does not? (March 2015, Oct 2020)
Cannizzaro Reaction:
Aldehydes that lack an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali. One molecule is oxidized to carboxylic acid salt and the other is reduced to an alcohol.
Reaction of Formaldehyde:
$2HCHO + \text{conc. } NaOH \xrightarrow{\Delta} CH_3OH \text{ (Methanol)} + HCOONa \text{ (Sod. formate)}$
Reasoning:
Formaldehyde ($HCHO$) does not possess any $\alpha$-carbon and thus no $\alpha$-hydrogen. Hence, it undergoes the Cannizzaro reaction. Acetaldehyde ($CH_3CHO$) possesses an $\alpha$-carbon with three $\alpha$-hydrogens, which makes it acidic enough to form an enolate ion in the presence of alkali, leading to Aldol condensation instead.
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